最小二乘法拟合椭圆附带matlab程序

1、最小二乘法拟合椭圆设平面任意位置椭圆方程为：x2 + Axy + By 2 + Cx + Dy + E = 0设 Pi ( xi ,yi )( i = 1,2, ,N) 为椭圆轮廓上的N( N 5 ) 个测量点， 依据最小二乘原理， 所拟合的目标函数为：N2F( A,B, C,D, E) = (x i2+ Axi yi + Byi2 + Cxi + Dyi+ E)i=1欲使 F 为最小，需使?F?F?F?F=?F?A=?D= 0?B?C?E由此可以得方程： xi2yi2 xi yi3 xi2yi xiyi2 xi yi xi3 yi34232A22 xiyi yi xiyi yi yi xiy

2、i xi2y xi y2 xi3 xiy xiB xi3iiC =-iD xiyi2 yi3 yi2 yi2 xi2 yi xi yi E xiyi yi2 xi yiN xi2解方程可以得到A，B， C， D， E 的值。根据椭圆的几何知识，可以计算出椭圆的五个参数：位置参数(,x0, y0 )以及形状参数( a, b) 。x0=2BC-ADA2-4By02D -AD=4BA2 -(2-D2+ 4BE-A2 )2 ACD-BCEa = 2-4B) (B -2B2)+ 1)( AA+ (1-2(ACD -2-D2+ 4BE-A2 )BCEb = 2-4B)(B +2(1 -B2)+ 1)( A

3、A += tan -1a2 - b 2Ba2 B -b 2附： MATLAB程序functionsemimajor_axis, semiminor_axis, x0, y0, phi = ellipse_fit(x, y)% Input:% x a vector of x measurements% y a vector of y measurements% Output:%semimajor_axis Magnitude of ellipse longer axis%semiminor_axis Magnitude of ellipse shorter axis%x0 x coordinat

4、e of ellipse center%y0 y coordinate of ellipse center%phi Angle of rotation in radians with respect to x-axis% explain% 2*b*x*y + c*y2 + 2*d*x + 2*f*y + g = -x2% M * p = bM = 2*x*y y2 2*x 2*y ones(size(x),%p = b c d e f gb = -x2.% p = pseudoinverse(M) * b.x = x(:);y = y(:);%Construct MM = 2*x.*y y.2

5、 2*x 2*y ones(size(x);% Multiply (-X.2) by pseudoinverse(M) e = M(-x.2);%Extract parameters from vector e a = 1;b = e(1); c = e(2); d = e(3); f = e(4); g = e(5);%Use Formulas from Mathworld to find semimajor_axis, semiminor_axis, x0, y0, and phi delta = b2-a*c;x0 = (c*d - b*f)/delta;y0 = (a*f - b*d)/delta;phi = 0.5 * acot(c-a)/(2*b);nom = 2 * (a*f2 + c*d2 + g*b2 - 2*b*d*f - a*c*g); s = sqrt(1 + (4*b2)/(a-c)2);a_prime = sqrt(nom/(delta* ( (c-a)*s -(c+a); b_prime = sqrt(nom/(delta* ( (a-c)*s -(c+a);semimajor_axis = max(a_prime, b_prime);semiminor_ax